按字母顺序排序 JSON（按特定元素）
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Sorting JSON (by specific element) alphabetically
提问by Alex Neigher
I have some JSON that is formatted like:
我有一些 JSON 格式如下：
places =[
{
"city":"Los Angeles",
"country":"USA",
},
{
"city":"Boston",
"country":"USA",
},
{
"city":"Chicago",
"country":"USA",
},
]
et cetera...
等等...
I am trying to sort this alphabetically BY CITYand am having trouble doing so. I believe the root of my issue seems to be determining the order of the characters (versus numbers). I've tried a simple:
我正在尝试按字母顺序按城市排序 ，但在这样做时遇到了麻烦。我相信我的问题的根源似乎是确定字符的顺序（相对于数字）。我试过一个简单的：
places.sort(function(a,b) {
return(a.city) - (b.customInfo.city);
});
yet, this subtraction doesnt know what to do. Can someone help me out?
然而，这个减法不知道该怎么做。有人可以帮我吗？
回答by Matti Virkkunen
Unfortunately there is no generic "compare" function in JavaScript to return a suitable value for sort(). I'd write a compareStrings function that uses comparison operators and then use it in the sort function.
不幸的是，JavaScript 中没有通用的“比较”函数来为 sort() 返回合适的值。我会编写一个使用比较运算符的 compareStrings 函数，然后在排序函数中使用它。
function compareStrings(a, b) {
// Assuming you want case-insensitive comparison
a = a.toLowerCase();
b = b.toLowerCase();
return (a < b) ? -1 : (a > b) ? 1 : 0;
}
places.sort(function(a, b) {
return compareStrings(a.city, b.city);
})
回答by Michael Geary
Matti's solution is correct, but you can write it more simply. You don't need the extra function call; you can put the logic directly in the sortcallback.
Matti 的解决方案是正确的，但您可以更简单地编写它。你不需要额外的函数调用；您可以将逻辑直接放在sort回调中。
For case-insensitive sorting:
对于不区分大小写的排序：
places.sort( function( a, b ) {
a = a.city.toLowerCase();
b = b.city.toLowerCase();
return a < b ? -1 : a > b ? 1 : 0;
});
For case-sensitive sorting:
对于区分大小写的排序：
places.sort( function( a, b ) {
return a.city < b.city ? -1 : a.city > b.city ? 1 : 0;
});
